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Chapter 9: Problem 554

Find the expected value and variance of a random variable, $$ \mathrm{Y}=\mathrm{a}_{1} \mathrm{X}_{1}+\mathrm{a}_{2} \mathrm{X}_{2}+\ldots \ldots+\mathrm{a}_{\mathrm{n}} \mathrm{X}_{\mathrm{n}} $$ where the \(\mathrm{X}_{\mathrm{i}}\) are independent and each have mean \(\mu\) and variance \(\sigma^{2}\). The \(a_{i}\) are constants.

Short Answer

Expert verified
The expected value and variance of the random variable Y are: Expected value: \(E[Y] = \mu(a_1 + a_2 + \cdots + a_n)\) Variance: \(Var(Y) = \sigma^2(a_1^2 + a_2^2 + \cdots + a_n^2)\)

Step by step solution

01

Write the random variable Y as a sum of its components

Define the random variable Y as the linear combination: \(Y = a_1X_1 + a_2X_2 + \cdots + a_nX_n\)
02

Find the expected value of Y

Use the linearity property of expectation, which states that \(E[aX + bY] = aE[X] + bE[Y]\) for any constants a and b and random variables X and Y. Apply this property to the given random variable Y: \(E[Y] = E[a_1X_1 + a_2X_2 + \cdots + a_nX_n]\) \(E[Y] = a_1E[X_1] + a_2E[X_2] + \cdots + a_nE[X_n]\) Since the mean of each random variable \(X_i\) is \(\mu\), we can rewrite the equation as: \(E[Y] = a_1\mu + a_2\mu + \cdots + a_n\mu = \mu(a_1 + a_2 + \cdots + a_n)\)
03

Find the variance of Y

Use the property of variance for independent random variables, which states that \(Var(aX + bY) = a^2Var(X) + b^2Var(Y)\) for any constants a and b and independent random variables X and Y. Apply this property to the given random variable Y: \(Var(Y) = Var(a_1X_1 + a_2X_2 + \cdots + a_nX_n)\) \(Var(Y) = a_1^2Var(X_1) + a_2^2Var(X_2) + \cdots + a_n^2Var(X_n)\) Since the variance of each random variable \(X_i\) is \(\sigma^2\), we can rewrite the equation as: \(Var(Y) = a_1^2\sigma^2 + a_2^2\sigma^2 + \cdots + a_n^2\sigma^2 = \sigma^2(a_1^2 + a_2^2 + \cdots + a_n^2)\)
04

Present the results

We have calculated the expected value and variance of the random variable Y as: Expected value: \(E[Y] = \mu(a_1 + a_2 + \cdots + a_n)\) Variance: \(Var(Y) = \sigma^2(a_1^2 + a_2^2 + \cdots + a_n^2)\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Random Variables
In probability and statistics, a random variable represents a quantitative feature whose value depends on the outcomes of a random phenomenon. For instance, tossing a coin results in two possible outcomes: heads or tails. If we assign a value such as 1 to heads and 0 to tails, we have created a random variable that interprets the outcome of the toss.

A random variable can be discrete, taking on specific isolated values like the toss of a die, or continuous, such as the temperature on a given day. Understanding random variables is crucial as they are foundational in defining and calculating statistics like means (or expected values), variances, and probabilities of different outcomes.
Linearity Property of Expectation
The linearity property of expectation is a fundamental concept in statistics that simplifies the calculation of the expected value for a combination of random variables. According to this property, the expected value of a sum (or linear combination) of random variables is equal to the sum of their individual expected values, regardless of whether they are independent or not.

Mathematically, for constants \(a\) and \(b\) and random variables \(X\) and \(Y\), this is expressed as:\
\[ E[aX + bY] = aE[X] + bE[Y] \]
This property is powerful because it allows us to perform calculations on the expected values of complex expressions by breaking them down into simpler parts.
Variance Properties
Variance measures how much a set of random variables deviates from their expected value, indicating the spread of a distribution. A key property of variance is that, unlike expectation, it is not a linear operation, but it does hold a special property for independent random variables.

Specifically, for any constants \(a\) and \(b\), and for independent random variables \(X\) and \(Y\), the variance of their linear combination can be expressed as:\
\[ Var(aX + bY) = a^2Var(X) + b^2Var(Y) \]
For independent variables, variances add up after being scaled by the square of the constants used in their linear combination, signifying that the overall variability increases with the square of the scaling factors. This property is extremely useful when handling variables that are affected by different scales of influence.
Independent Random Variables
Random variables are independent if the occurrence of one does not affect the probability distribution of the others. This concept is essential when dealing with multiple random processes, as independence simplifies calculations and allows the use of the variance property mentioned earlier.

Independence is a strong assumption that may not hold in all cases, especially in complex systems where variables often interact. However, when confirmed, it enables us to treat individual random variables in a vacuum for various predictive and analytical calculations, making the process more manageable.
Linear Combinations of Random Variables
A linear combination of random variables occurs when these variables are summed together, each being multiplied by a constant. This is a common operation when constructing new variables from existing ones, such as creating an index or score.

The importance of a linear combination lies in its ability to integrate distinct random processes into a single measure, which can then be analyzed using the expectations and variances of the original variables. In the context of the exercise mentioned, understanding that the expected value of a linear combination is the weighted sum of the expected values, and the variance is the sum of the squared weights times the variances, provided the variables are independent, greatly simplifies the process of assessing the behavior of complex models.

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Most popular questions from this chapter

Find the probability that a person flipping a balanced coin requires four tosses to get a head.

Suppose we want to compare 2 treatments for curing acne (pimples). Suppose, too, that for practical reasons we are obliged to use a presenting sample of patients. We might then decide to alternate the 2 treatments strictly according to the order in which the patients arrive \((\mathrm{A}, \mathrm{B}, \mathrm{A}, \mathrm{B}\), and so on). Let us agree to measure the cure in terms of weeks to reach \(90 \%\) improvement (this may prove more satisfactory than awaiting \(100 \%\) cure, for some patients, may not be completely cured by either treatment, and many patients might not report back for review when they are completely cured). This design would ordinarily call for Wilcoxon's Sum of Ranks Test, but there is one more thing to be considered: severity of the disease. For it could happen that a disproportionate number of mild cases might end up, purely by chance, in one of the treatment groups, which could bias the results in favor of this group, even if there was no difference between the 2 treatments. It would clearly be better to compare the 2 treatments on comparable cases, and this can be done by stratifying the samples. Suppose we decide to group all patients into one or other of 4 categories \- mild, moderate, severe, and very severe. Then all the mild cases would be given the 2 treatments alternatively and likewise with the other groups. Given the results tabulated below (in order of size, not of their actual occurrence), is the evidence sufficient to say that one treatment is better than the other? $$ \begin{array}{|c|c|c|} \hline \text { Category } & \begin{array}{c} \text { Treatment A } \\ \text { Weeks } \end{array} & \begin{array}{c} \text { Treatment B } \\ \text { Weeks } \end{array} \\ \hline \text { (I) Mild } & 2 & 2 \\ & 3 & 4 \\ \hline \text { (II) Moderate } & 3 & 4 \\ & 5 & 6 \\ & 6 & 7 \\ & 10 & 9 \\ \hline \text { (III) Severe } & 6 & 9 \\ & 8 & 14 \\ & 11 & 14 \\ \hline \text { (IV) Very severe } & 8 & 12 \\ & 10 & 14 \\ & 11 & 15 \\ \hline \end{array} $$

Flapjack Computers is interested in developing a new tape drive for a proposed new computer. Flapjack does not have research personnel available to develop the new drive itself and so is going to subcontract the development to an independent research firm. Flapjack has set a fee of \(\$ 250,000\) for developing the new tape drive and has asked for bids from various research firms. The bid is to be awarded not on the basis of price (set at \(\$ 250,000\) ) but on the basis of both the technical plan shown in the bid and the firm's reputation. Dyna Research Institute is considering submitting a proposal (i.e., a bid) to Flapjack to develop the new tape drive. Dyna Research Management estimated that it would cost about \(\$ 50,000\) to prepare a proposal; further they estimated that the chances were about \(50-50\) that they would be awarded the contract. There was a major concern among Dyna Research engineers concerning exactly how they would develop the tape drive if awarded the contract. There were three alternative approaches that could be tried. One involved the use of certain electronic components. The engineers estimated that it would cost only \(\$ 50,000\) to develop a prototype of the tape drive using the electronic approach, but that there was only a 50 percent chance that the prototype would be satisfactory. A second approach involved the use of certain magnetic apparatus. The cost of developing a prototype using this approach would cost \(\$ 80,000\) with 70 percent chance of success. Finally, there was a mechanical approach with cost of \(\$ 120,000\), but the engineers were certain of success. Dyna Research could have sufficient time to try only two approaches. Thus, if either the magnetic or the electronic approach tried and failed, the second attempt would have to use the mechanical approach in order to guarantee a successful prototype. The management of Dyna Research was uncertain how to take all this information into account in making the immediate decision-whether to spend \(\$ 50,000\) to develop a proposal for Flapjack. Can you help?

A sample of size 49 yielded the values \(\underline{x}=87.3\) and \(s^{2}=162 .\) Test the hypothesis that \(\mu=95\) versus the alternative that it is less. Let \(\alpha=.01\).

Given a normal population with \(\mu=25\) and \(\sigma=5\), find the probability that an assumed value of the variable will fall in the interval 20 to 30 .
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